HDU 1394 minimum inversion number

Link: HDU 1394 minimum inversion number

This is an extension of the smallest reverse order. You can use a tree array. For a sequence of $ N $ numbers $ A $, the number of $ I $ ($ I \ in [0, n) $) the number of reverse orders $ r_ I $ can be expressed as its badge $ I $ minus the number before it and not greater than its number. For example, a [8] = 4 for the number in sequence a =. The number of reverse orders $ r_8 = 8-3 = 5 $, and the second 3 indicates three numbers that are earlier than it and smaller than it: {1, 3, 0 }. Then we can obtain the Reverse Order Number Formula for the number of $ I $:

\ Begin {equation} r_ I = I-\ sum _ {0 \ Leq j <I and a_j \ Leq a_ I} 1 \ end {equation}

For the entire sequence $ A $ has

\ Begin {equation} r = \ sum _ {0 \ Leq I <n} r_ I \ end {equation}

You can use the sum (I) function of the first I item of the tree array to quickly calculate $ \ sum _ {0 \ Leq j <I and a_j \ Leq a_ I} 1 $, and use add (I + 1) to count the number.

After obtaining the reverse Number of the initial sequence, you can easily calculate the reverse Number of the transformed sequence.

If you delete the first number $ A_1 $, the number of reverse orders is reduced by $ A_1 $.

Put the first number at the end of the series, then the number of Reverse Order will increase $ n-A_1-1 $

In this way, you can try all the transformation methods and finally obtain the minimum value.

The Code is as follows:

 1 #include <cstdlib> 2 #include <cstdio> 3 #include <iostream> 4 #include <cstring> 5 #define MAXN  5005 6 using namespace std; 7 int bit[MAXN]; 8 int arr[MAXN]; 9 int n;10 int lowbit(int x)11 {12     return x&(-x);13 }14 int sum(int x)15 {16     int ans = 0;17     while( x > 0 )18     {19         ans += bit[x];20         x -= lowbit(x);21     }22     return ans;23 }24 void add(int l, int x)25 {26     while( l <= n )27     {28         bit[l] += x;29         l += lowbit(l);30     }31 }32 int main(int argc, char *argv[])33 {34     while( scanf("%d", &n) != EOF )35     {36         int ans = 0;37         memset(bit, 0, sizeof(bit));38         for( int i = 0 ; i < n ; i++ )39         {40             scanf("%d", &arr[i]);41             ans += i - sum(arr[i]+1);42             add(arr[i]+1, 1);43         }44         int mi = ans;45         for( int i = 0 ; i < n ; i++ )46         {47             ans = ans - arr[i] + n - arr[i] -1;48             mi = ans > mi ? mi : ans;49         }50         printf("%d\n", mi);51     }52 }

 

HDU 1394 minimum inversion number